text:
How can I avoid downdrafts and icing when crossing the Alps in a small airplane?
let's say I need to fly from LSZL to LSZP crossing the Alps with a small airplane (C150, Piper Tomahawk,...). How many times a year I could expect to find the suitable weather conditions to do that (both VFR and IFR)?
The main issues I have identified are:

Strong downdrafts near the base of the Alps (-2000fpm). A small plane has a service ceiling of 13000ft: would it be possible to climb high enough to avoid this phenomenon?
Icing: I expect the OAT to be below zero all year, moreover ice could build up even if I'm outside the clouds. Normally small planes are not equipped with boots, or de-icing fluids systems. How can this problem be handled?

Airframe icing requires visible moisture, so if you stay clear of it, you should not pick up airframe icing.

Welcome to the site! If you don't get any useful answers here, you could look for help from a local flying or gliding club in that area. Local knowledge is really helpful for both weather questions and mountain flying, especially if you plan to fly through the mountains rather than over them.

If you're planning on (even possibly) exceeding 10k feet, BRING OXYGEN.

Yes, a low powered trainer can fly through tall mountains.
 It is not possible to say how often the flight may be made safely as it depends on specific weather conditions across your route of flight.
This is the same as saying a small sailboat, like a sabot, can sail to the Azores.  It is possible, and might even be safe with proper planning and good weather, but you must take care to plan the trip well every time you make it.
The number one concern I would have flying through the Alps (...or Rocky Mountains, or Sierra Nevadas...) is wind, temperature being a very close second.  A low powered trainer is pretty much at the mercy of the wind much above 5,000' and is absolutely helpless above 10,000'.  A very tame downdraft of only a few hundred feet per minute will exceed the plane's best rate of climb and could effortlessly push you into a ridge or, if you're lucky, a valley or canyon from which you may not be able to climb out.  I use the example of the Cessna 150 in Meteor Crater (elevation only 5,700' in Arizona, USA) of what can happen when an under-powered aircraft runs afoul of nature at high density altitudes.  Ice is not really an issue unless you are willing to fly through visible moisture...so...don't fly though clouds, mist, fog, etc.
I earned many of my hours flying through the Rockies in piston aircraft. The vast majority of my time in the Rockies was flying freight in cabin-class Twin Cessnas with turbocharged, geared, high displacement Continental engines and de-ice kits (a capable airplane).  Even in these planes it was common to encounter downdrafts that far exceeded aircraft performance even when using takeoff power and when lightly loaded.  The turboprops that I flew later on would fare better in that they would hold altitude (maybe even climb) at Vy and max power while carrying a load.  A Cessna 150/152, Tomahawk, or anything else with an O-235 will not do well in breezy mountain conditions.  The large mountain ranges in the US are dotted with light plane wreckage due to unfavorable weather and wind conditions.
There are things you can do to more safely traverse high terrain in a low powered plane.  I learned a few of these the hard way...
1.) Learn proper ridge crossing techniques.  There is a well defined "right way" to approach ridges if you have any doubt as to the wind conditions across the ridge.  I provide FAA documentation only because it is what I am most familiar with.
2.) Set very high weather minimums for trips through mountains.  Twice the standard requirements for VFR may be a good start, perhaps even higher.  Adhere to them during your flight.
3.) Identify alternate airports.  Know the status of the alternates (plowed?), the services provided (fuel?), and their local weather forecasts before you depart.  Take time to note distance and direction to the nearest alternates throughout your flight.  The weather can turn quickly and simply turning around and flying home is typically not an option.  Have frequencies and airport diagrams close at hand should you need an alternate.
4.) Have two realistic emergency landing spots picked out at all times.  You don't need have spots that will allow for a complete rollout, just spots that will permit you to dissipate enough energy to survive the impact. I've built the habit to scan sky, then ground, then sky when flying over rough terrain.
Landing in a field and damaging the plane may be wiser than continuing into worsening conditions.  This is a difficult realization to accept, but attempting to fly through worsening visibility, breezy conditions, and any amount of visible precipitation may kill you while landing in a field may just kill the plane.
5.) Universal advice, but particularly relevant here...if forced to land, fly all the way through the crash.  Pilots that fly their planes through a crash survive.  Pilots who become passengers do not.
If this sounds like a lot of work, it is! If this sounds overly dramatic, it isn't.  The low powered trainer was not really designed for this type of flight.  You need to take additional measures to make sure that the conditions along the route conform to the aircraft's rather severe performance limitations.
You also need to accept that the plane you describe may be quickly and thoroughly outclassed by mountain weather, weather that even a moderately more capable aircraft (C172-180hp, Piper Archer, ...) would not have a serious problem with.
This flight is doable can be done safely if planned well and flown in favorable conditions.

A good source of air currents around the alps will be from the gliding, hangliding and paraglider pilots who fly the area regularly and depend on these air currents to stay airborne.
In general, unless you are planning an instrument rating flight through clouds, you can expect to see significant numbers of recreational pilots centred around most tourist areas in the alps, with the possibility of a cross country pilot almost anywhere.
In general terms, the air is more likely to be rising on the sunny side of the valley. Keeping an eye on the cumulus clouds will give you an idea on the relative strength and location of significant areas of updrafts. Any clouds bubbling up like Marge Simpsons hair style indicate extreme up drafts - other areas are likely to have strong down drafts.
Late afternoon, when the sun generated thermal activity stops, expect large areas of sinking air and 180 degree wind speed direction changes in valleys. Narrow valleys will have stronger valley winds.
Your principle problem is that this section of the Alps is not blessed with North / South running valleys. The pass roads are some of the highest in the Alps at around 7,000' to 8000' typically only open May - October. Road transport between Switzerland and Italy is primarily by tunnels.
Have a look at flying towards Geneva, then flying south crossing around Col du Grand St Bernard's into Italy as one possible route.


text:
Why do commercial pilots always engage reverse thrust even if it is not necessary?
Most of the time the usage of reverse thrust is not needed. I wonder if it's an authority (e.g. FAA) requirement to land safely.
Obviously, in bad weather conditions or when the runway is contaminated, the brakes might not work as effectively and thus R/T may be required to land safely. However, in most of other situations the aircraft can brake OK. Given that it would probably be more efficient to land without using R/T, why don't airlines routinely do so?

Related, maybe duplicate: For large jets, what is the primary means of slowing down after landing? Also: How does thrust reversal work on a jet engine, and when should it be used?

Related, and should answer the question: http://aviation.stackexchange.com/questions/24629/why-do-modern-jetliners-provide-thrust-reversers-at-all-in-their-design-decision

NO!!!! This is NOT a duplicate of those questions! Those are "engineering" questions about reverse thrust and the other ways to stop the aircraft; this question is asking about POLICY and (ultimately) the HUMAN FACTORS behind that policy.  Not the same question, and not the same answers!

The question is, "why do comercial pilot always engage R/T"?  That's a matter of policy.  The point that it is often unnecessary is entirely valid, and at most the links to the other questions simply support that statement.  WITHIN the body of the question, the (essentially rhetorical) question "wouldn't it be more efficient..." is answered in those questions, but the POINT of what's being asked is, given all of that, why do pilots always use R/T anyway?  And THAT question isn't addressed elsewhere as far as I can tell.  Everything else simply supports that "why" question.

The correct answer is, habit patterns.
If your habit, as a pilot, is to ALWAYS select reverse thrust, you'll do so reliably even when the weather is bad, you're fighting gusty crosswinds, and there is whatever else going on. On the other hand, if you sometimes do & sometimes don't select reverse thrust, that habit pattern isn't there to back you up when you have the proverbial "dark & stormy night" and getting into reverse thrust right away may be the difference between staying on the runway or running off the end.
For example, if I'm landing a fairly light aircraft on a 12,000' dry runway in DEN with a little headwind, it is entirely possible to leave the thrust reversers stowed, roll out with the speed brakes deployed, tap the brakes slightly, and exit on the last high speed taxiway at a normal taxi speed.  And I'd burn less fuel and make less noise doing that.  But at my carrier, I'm not allowed to, because it's considered that the habit pattern of always deploying the T/R is so important that it's worth it to burn that extra gas even when it isn't strictly necessary, so that on the day when it's vital, the habit pattern to grab the T/R's right after touchdown is strong.
I don't think that this is an OpSpec (i.e. FAA) requirement, at least for the aircraft that I fly, but I'm quite certain that it is a requirement in our books, so my supposition is that it's a company requirement.  If that assumption is correct, then other companies might choose to do differently (and I recall one early morning arrival into Frankfort on a Lufthansa A-320 that used NO reverse thrust, but some heavy braking instead -- presumably to reduce noise at that hour of the morning).  The fact that safety is so important, though, probably drives plenty of operators to make the same choice that our carrier does: build the habit of ALWAYS using reverse thrust -- even though that means sometimes using it when it isn't necessary.

I very much agree that deploying the reversers as soon as possble is a good habit pattern. However, perhaps it should be noted that because you deploy them doesn't mean you have to use other than idle thrust with them deployed. Deploying bucket style reversers but staying at idle power gets you a fair amount of drag while the airplane is still at high speed right after touchdown. This is what we typically did at the two 747 carriers I worked at when we had a long runway and had no need to turn off earlier than that allowed.

Excellent points.  Our policy is to deploy the reversers to at least Detent 2, which is beyond idle power, and our engines use sleeves rather than buckets.  Again, desired habit patterns.  Of course, they don't say how long we have to stay in reverse, so you could coast down to the last highspeed exit after stowing them.  Each operator writes their own rules in that regard, and I'm not claiming that ours is optimal -- just mentioning what it is.  The buckets-out+idle sounds like an excellent solution as well.

Thank you for such a good answer, that was what I was trying to know @RalphJ

Also perhaps worth noting that the extra gas to deploy TR's, even over a few hundred landings, is still a heck of a lot cheaper than writing off an airframe on that one time you do dump your brand new A380 off the end of the runway.

Could it also be to protect against a break failure? If you suddenly lose break power, the last thing you want is a delay while the thrust reverser deploy and activate. By deploying them in advance, you have them armed and ready if needed.

How do you know it wasn't necessary?

Maybe there are planes in the pattern ready to land immediately after you. Delaying clearing the runway could effect the safety and schedules of dozens of other aircraft.
Maybe your brakes or tires are going to fail partway into your braking leaving you no time to use reverse thrust
maybe a portion of the runway IS contaminated. The only way to avoid it is to slow as soon  can. The longer you are at high speed the more likely a problem like this will effect you
Something (another plane or animal) might encroach on the runway while you are slowing down lackadaisically
Wind or other weather phenomenon might effect your high speed landing roll. The period of time when a plane is near flight speeds while touching the ground is one of the most dangerous times in a flight. Many bad things can happen while "fast taxiing"
Missing an earlier taxiway might result in a longer taxi that's even more expensive than a little thrust reversal.

The danger level is so elevated, there is just no reason good enough to maintain high speed while on the ground, other than when you intend to take off.

How can we say that it's not necessary?
I'd say that the reason I use reverse more then brakes is that it's my airline procedure and Airbus says that too!
Use of brakes to slow down airplane after high speed landing is not good for tyre's health. So what we do normally is slow down the airplane by thrust reversers till 40-60 knots and then we use brakes.

This doesn't seem to add anything additional from the two answers that were already posted

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Is it allowed for state aircraft to fly within RVSM airspace?
If the answer is no, then it will fly at FL280 or 430? Or it will fly within RVSM airspace with vertical separation of 2000?

Would you mind adding some context? Those us not familiar with state aircraft might be having a hard time understanding why this is an issue. Are state aircraft generally incapable of RVSM?

If the state aircraft is certified for RVSM then in can fly in RVSM airspace.

In the usa, ATC guidance is as follows (JO 7110.65 2–1–29a1):

Ensure non-RVSM aircraft are not permitted in RVSM airspace unless they meet the criteria of excepted aircraft and are previously approved by the operations supervisor/CIC. The following aircraft are excepted: DoD, DoD-certified aircraft operated by NASA (T38, F15, F18, WB57, S3, and U2 aircraft only), MEDEVAC, manufacturer aircraft being flown for development/certification, and Foreign State aircraft. These exceptions are accommodated on a workload or traffic-permitting basis.

The Chicago Convention, which established ICAO, does not apply to State or military aircraft. This includes the restriction on aircraft that can operate in RVSM airspace. From an ICAO working paper:

The Convention on International Civil Aviation (Chicago Convention) and its Annexes provide the legal
and operational framework for Member States, inter alia, to operate aircraft in airspace where a reduced
vertical separation minimum (RVSM) is applied. The Convention shall be applicable only to civil
aircraft, and shall not be applicable to State aircraft.

States are free to apply regulations on their own State aircraft, which may or may not permit them to operate in RVSM airspace.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Is there any instrument on board to assure horizontal collision avoidance?
I know that TCAS are used to get 'Traffic Alerts' as well as 'Resolution Advisories'.
But an RA seems to only provide instructions to avoid vertical collisions. How about the horizontal separation, how do we manage that ?

You seem to misunderstand how TCAS works. Have a look at How does TCAS work? While an RA only gives vertical instructions, it avoids all collisions, including horizontal ones.

Does this answer your question? How does TCAS work?

Not a dupe, as this question is asking about equipment other than TCAS. The fact that there isn't any such equipment doesn't make it a bad question.

A TCAS Resolution Advisory directs a vertical vector such as climbs and descents, but it is aware of both the horizontal and vertical position of other aircraft. After all, a collision is a collision, and the distinction between "horizontal collision" and "vertical collision" is a little artificial. TCAS uses vertical vectors to avoid any collision, whether or not there was any vertical component in the original collision-course vectors.
As far as systems that direct horizontal (ie heading) changes to avert a collision, that's ATC's role, and of course navigation systems to keep everybody "where you're supposed to be" will prevent all sorts of potential problems. But no on-board systems direct heading changes in response to another aircraft in the same way that TCAS does in the vertical dimension.

To avoid colliding, two aircraft need horizontal separation or vertical separation, not both. If two aircraft are on a collision course, this means that their tracks and altitudes are converging. An RA will provide altitude separation. Their tracks will still cross but this is fine and normal.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
On coaxial helicopters, how exactly does the yaw mechanism work?

This question asks how coaxial helicopters achieve yaw. The accepted answer correctly states that this is done by increasing pitch on one rotor while decreasing pitch on the other rotor, which leaves lift the same and creates a yawing moment.
The post became active again on 20 July, when @Andy Ewen asked how the mechanism works exactly:

..a lot of the answers are stating the obvious but fail to explain HOW. Yes a torque differential between the rotors will effect a yaw but the real question is how is this achieved?

The post was deleted and not entered by OP as a separate question, hence this question. How is this achieved, exactly?

@ymb1 It's a boggle-copter, boggles the mind.


text:
How is take-off power defined?
In a recent post it could be read that by definition, turbofans do not produce power in the static case (at velocity zero). The statement could be a bit confusing since turbofans and propeller engines are at their highest power setting just before the take-off roll commences.
What is the referenced definition that results in zero power for turbofans at zero speed, regardless of the throttle setting?

Let me get this straight, you think that at full throttle an engine is producing no power if the takeoff roll has not yet started?  I am not sure where you get such an idea...

@Koyovis can you link to the post?

Is it a confusion between power and work done?

You're asking about this answer by Peter Kämpf, in which he writes (emphasis added):

Now for turbofans. Here we have thrust which needs to be converted into power first by multiplying it with flight speed. It would be nonsensical to compare the static case – here by definition turbofans do not produce power.

You ask, "What is the referenced definition that results in zero power for turbofans at zero speed, regardless of the throttle setting?" The referenced definition is the sentence in bold above. Peter is defining power as thrust multiplied by flight speed.
Now, let's back up a little bit. Why would Peter define power like that?
For propeller engines (and car engines), defining the "output power" of an engine is easy: it's the power being transmitted through the drive shaft (which is the product of the torque the engine is exerting on the drive shaft, and its rate of rotation). I'm going to call this definition "shaft power."
A jet engine, however, doesn't have a drive shaft, so we are unable to use this definition. We're forced to choose a different definition of power, and so Peter chooses to use the product of thrust and flight speed. I'm going to call this definition "ultimate power". (This generalizes to any vehicle; we can define "ultimate power" as the forward force on the vehicle produced by its engine, times the vehicle's forward speed. Equivalently, "ultimate power" is the rate at which the engine adds kinetic energy to the vehicle.)
However, Peter also points that power as defined this way (ultimate power) can't be directly compared with the other definition (shaft power), at least not when the vehicle is stationary. This is because the "ultimate power" for a stationary vehicle is always zero, whereas the "shaft power" for a stationary vehicle is generally much greater than zero.
It may seem counterintuitive that the "ultimate power" for a stationary vehicle is always zero, but it is. Kinetic energy is proportional to the square of speed, which means that if an object is traveling extremely slowly (as in the first few nanoseconds after it starts moving), it gains kinetic energy at an extremely slow rate (on the order of nanowatts).

Yes this definition would make sense for a pure turbojet. However the text is about a turbofan, which does have a drive shaft for the fan, which produces most of the thrust. The "ultimate power" for any vehicle at standstill is zero, also for turboprops, cars etc. I'm wondering what makes the definition valid for turbofans and not for turboprops.

@Koyovis Uhhhh, good point. I'm not sure if Peter overlooked that or if he deliberately decided not to define power based on the turbofan's drive shaft.

Why not define power as thrust produced?

@MichaelHall Well thrust would be force, not power.

Ok then, you are equating power to work performed.  I'm fine with that definition in a purely academic discussion, but it doesn't actually provide any useful insight.


text:
What are the design parameters for airliner safety?
Airliners designed to carry multiple airfare paying passengers are regulated by CFR 14 Part 25, and by EASA CS-25. The design regulations are such that the probability of a catastrophic failure with multiple deaths is extremely remote or extremely improbable.
How is that defined? Can it be quantified at all?

Extremely improbable is defined as 10$^{-9}$ per flight hour, or one in a billion flight hours. For instance, from EASA CS-25 Detail Design Requirements 4.3:

The actuation system minimum requirements should also be continuously met after any combination of failures not shown to be extremely improbable (occurrence less than 10$^{-9}$ per flight hour).

For aircraft systems and structures, that means we need to consider the following:

What is the probability that failure of function causes a catastrophe?
What is the exposure time to the failure condition?
What is the failure rate of the hardware?

Flight control systems are flight critical: without it functioning, we may assume that a catastrophe will occur. NASA-TM-72860 depicts the accumulated failure analysis factors for flight control systems like this:

If we take the cruise autopilot as an example:

When it fails, it just means that the pilot will hand-fly the aircraft, and the probability of its failure causing a catastrophe is set to one in a million flying hours. Pretty remote.
During cruise, the autopilot is on all the time. Exposure is pretty close to 100% (= 1)
That requires the cruise autopilot system to have a reliability of 1 failure per thousand flying hours.

The reliability demand is for system functionality, including sensors, processors, electronics, power supply etc. One way to meet functionality demand is to duplicate the system, so that if one system fails, one or more remaining systems will still carry out the function. Not really necessary for the cruise autopilot 10$^{-3}$ demand, but for instance the auto land system must be 1000 times more reliable: it is only functional during landing, but the probability of function loss causing a catastrophe during landing can be quite high due to proximity of the ground and very short reaction times.
That is why the CAT III auto land systems are up to triple redundant - the complete system is duplicated three times - if we only need one functional system and we have three, the functionality failure rate is $f(sys)^3$ Each individual system can then have a failure rate of 10$^{-2}$ per flight hour.
To put these numbers into perspective: the average B777 now flies 12 hours per day, that is 4,380 hours per year. It would have to fly for 228,000 years in order to experience a catastrophic failure with a chance of 1 per 10$^{-9}$ flying hours.

EDIT
Some very good questions from the comments. I'm answering them here, since all comments may be removed once the tally is 20 or more.
@Michael But how are those probabilities determined? Verification through testing gets harder with increasing safety: in order to get any statistical relevance, one would have to subject a safety critical system to 10 billion operational hours and track if there are indeed 10 functional failures. Operational records of all aircraft throughout history provide a frame of reference. Most design changes in aircraft are incremental for this reason, build up operational history of the aircraft first. Design parameter restrictions for military aircraft are much lower, and that is where most revolutionary design changes come from.
@James S:It appears all of these p(Catastrophes) are in isolation. Is there any attempt to calculate chance of catastrophe when there are multiple failures, or chance of failure when one item has already failed?
Yes:

Common mode failures, for instance EMD after an atomic blast that stops all electronics from functioning.
Hidden failures from back-up systems, or systems not in use at the moment. Redundancy is always better because it is always known that the system is operating.
Cascade failures, where one blade tears through all three hydraulic systems.

All of these failure modes may not result in more than 10$^{-9}$ per flight hour. For instance the Gust Load Alleviation: it is only required during an actual gust, but the status of the system since the last test or operation is unknown. Therefore the system reliability used in these failure calculations will have to include all possible failures since the last time the equipment was tested.
@Michael Kjörling Wikipedia claims that 1,526 Boeing 777 aircraft (all models) have been delivered. ... There have been six hull losses, minus one due to criminal act, or one non-criminal hull loss per about 31 million (about 10^7.5) flight hours. The estimates seem to be off by at least one order of magnitude. A hull loss without multiple deaths is not defined as a catastrophe. The design parameters only account for systems functionality and structural integrity, not acts of war. ...error conditions arising from system failures which did not result in a hull loss. are not defined as catastrophic, if a procedural change can fix the situation then the reliability requirements are less strict. The 10$^{-9}$ is a lower limit design parameter: for the B-777 case, divide the actual catastrophes as defined by authorities by the actual flying hours of the fleet, and it becomes apparent how excellent the safety numbers are. But they only start to gain statistical accuracy now, up to 5 - 10 billion flying hours accidents could all be flukes.
Then again: multiple similar catastrophic failures of a new fleet within a short time period may not be coincidental.

But how are those probabilities determined? I can see that it would make sense for more or less random, statistical processes (e.g. a cosmic ray hitting and flipping a memory bit in your flight computer). But how do you determine the probability of a turbine blade fracturing and consequently smashing through all 3 redundant hydraulic systems?

It appears all of these p(Catastrophes) are in isolation. Is there any attempt to calculate chance of catastrophe when there are multiple failures, or chance of failure when one item has already failed?

Wikipedia claims that 1,526 Boeing 777 aircraft (all models) have been delivered. That would be about 6.7 million flight hours per year. Let's play generous and say that all of those were delivered in the year of its first flight, 1994, and still fly (obviously neither being the case); that brings us to some 154 million flight hours total. There have been six hull losses, minus one due to criminal act, or one non-criminal hull loss per about 31 million (about 10^7.5) flight hours. The estimates seem to be off by at least one order of magnitude.

@MichaelKjörling Only 1 of those hull losses that also had fatalities might have been due to a failure of the system, Malaysia Airlines Flight 17, all the others were either due to human error or had no fatalities

@MarkOmo Certainly fair enough about discounting hull losses which were caused by human/pilot error. I'm not sure I'd discount hull losses with no fatalities from consideration regarding error probability, though. Also note that I considered only hull losses, ignoring the (likely far more common) error conditions arising from system failures which did not result in a hull loss. Accounting for this would require a far more detailed study, data for which might not even be available, but I wouldn't be the least bit surprised if, to a first order approximation, they roughly cancel each other out.

@MichaelKjörling the 10^-9 number is for mutable fatality incidents if the hull was a total loss with no injuries the design did its job

@Michael The processes that manufacturers use in the development of their aircraft are based primarily on two SAE Aerospace Recommended Practices (ARP), ARP4754A Guidelines For Development Of Civil Aircraft and Systems and ARP4761 Guidelines and Methods for Conducting the Safety Assessment Process on Civil Airborne Systems and Equipment. Together they provide a way to identify safety needs, specify requirements and show compliance to section 25.1309.

@Gerry Indeed, plus for instance AC 25.571 for statistical design criteria for structural integrity during gusts.

@MarkOmo: Did you mean Malaysia Airlines Flight 370?  MH17 crashed due to being hit by a surface-to-air missile, which I'm pretty sure does not count as being caused by a failure of any aircraft system.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Why do tailplanes provide down force, if not for longitudinal static stability?
We can see that most conventional aircraft have their horizontal tail surfaces arranged to provide a down force. This question asks if that is required for static longitudinal stability, and the answer is: not necessarily.
But if static stability is not the reason, what is? Or is it just a meme indeed?

Simple answer: To provide enough static stability and a wider center of gravity range.
The lift on the tail will become negative if more stability is desired than what is achievable with no load on the tail. Another reason is a forward center of gravity location due to loading and securing the cargo on board or an uneven emptying of fuel tanks.
For usability, the center of gravity of every aircraft is allowed to vary, within clearly specified bounds. The forward limit is usually determined by:

Maximum stick forces in a pull-up maneuver (that used to be specified in MIL-F-8785C §3.2.3.5, for example). During the certification flights of the Do-228 for the CAA, the pilot flew a dive with full forward cg location and let the trim run away completely, waiting too long to take corrective action. His last words were "help me on the stick!".
Maximum stick forces over load factor. If the pilot needs to pull positive gs, the aircraft should not fight back too much.
Sufficient control authority in ground effect. When flying close to the ground, the effectiveness of the elevator is reduced. In order to have enough pitch control authority to rotate and lift off, the cg must not be too far forward. Neglecting this led to a kludgy fix during the development of the F-18, the rudder toe-in.
Speed stability: The aircraft should maintain the trimmed speed, even in gusty weather, without requiring constant pilot inputs. This is the main reason for negative tail loads: Comfort.

I agree with your last point, about comfort (aka, sufficient positive static stability to allow the aircraft to be manageable when flown manually), being the main reason for this. You are clearly more technically conversant than I. Do you know whether it is even possible to design an aircraft with tail-mounted pitch control so that at cruise, there is positive lift on the tail surface, yet there is sufficient static margin to allow reasonable range of CG, and not create negative stability conditions at high (close to Clmax) AOA? Was any aircraft ever so designed?

@CharlesBretana: Static stability doesn't change with AoA, only tail load does in planes with wing camber. Modern Airbus planes should have neutral to slightly positive tail loads when the tail tank is used correctly (I have spoken with pilots who refused to use it). Stability is ensured by the autopilot which is even required for comfort with negative tail loads at high altitude due to the low pitch damping there. Note that the tail tank is only used to shift the cg back, so cargo can be placed in a wider cg range without affecting performance.

I don't mean to start another extended discussion, but what I am talking about is the forward motion of the aerodynamic center of any airfoil that occurs as the boundary layer separates at the trailing edge as AoA increases near the maximum (stall) AoA. Does not this forward motion caused by trailing edge boundary layer separation cause a decrease in static margin?

@CharlesBretana: Once separation starts, we leave the linear range of aerodynamics and all simplifications are no longer valid. Now the pressure over the separated area is lower, such that the center of pressure moves back with the onset of trailing edge separation. This actually stabilizes the aircraft! With full separation pressure is roughly constant over chord and the center of pressure is nearer 50%, not 25% as in attached subsonic flow.

From Wikipedia article on Center of Pressure (https://en.wikipedia.org/wiki/Center_of_pressure_(fluid_mechanics)#cite_note-10 ), "For a conventionally cambered airfoil, the center of pressure lies a little behind the quarter-chord point at maximum lift coefficient (large angle of attack), but as lift coefficient reduces (angle of attack reduces) the center of pressure moves toward the rear". So as AOA increases, it moves forwards. It moves forwards because the boundary layer flow separates from the airfoil at the rear trailing edge and this separation propagates forwards as AOA increases.

This phenomenon occurs in all conventionally cambered airfoils. The center of pressure remains fixed at the 25% MAC point only in symmetrically cambered airfoils, which are the exception, and not the norm.

But the article also states that the Aerodynamic center does not move as AOA changes. This surprised me, If this is the case (I Have no reason to disbelieve it) I do not understand it. Can you explain how this can be? if the center of pressure moves, how can the point through which the pitch moment acts remain constant?  What is the difference between the center of Pressure and the Aerodynamic Center?  see https://aviation.stackexchange.com/questions/19388/what-is-the-difference-between-centre-of-pressure-aerodynamic-centre-and-neutral

@CharlesBretana: Please keep linear aerodynamics (no separation) and nonlinear aerodynamics (separation included) apart. 95% of what you read implicitly concerns linear aerodynamics, without stating so explicitly. Next, keep the aerodynamic center/neutral point and center of pressure apart. Explained here, too.

@CharlesBretana: The forward movement of the center of pressure has nothing to do with separation, but comes purely from camber effects. Angle of attack produces a lift component at 25% (without separation!) which becomes stronger the more AoA increases. Camber adds another lift component with a more rearward center of pressure which is independent of AoA. Both combine, and the AoA-dependent part becomes stronger with higher AoA. Only this makes the center of pressure move forward in linear flow.

So what is the change in pilot comfort and stability if a taiplane is providing an upforce on a regular plane (not a canard etc)?

@Konrad Lower control forces for maneuvering, but it becomes unwise to let go of the stick in gusty weather. You need to watch your speed more closely with lower stability, too.

Because the safe situation is: a gravitational moment nose down, trimmed with an aerodynamic moment nose up. In all flight conditions. The critical situation is not cruise, but the conditions at low speed.

At TO and landing the Angle of Attack is high: a nose up attitude, commanded by a downward tail force. Velocity is low, so the large force is achieved by a large upward deflection of the elevator and/or a negative stabiliser incidence. The AoA range is from around zero in cruise, to 15-20 deg nose up: the required elevator range is from zero deflection to large up deflection (plus superimposed manoeuvre requirements down & up). The average required force is downwards.
You don't ever want the horizontal control surface to stall, but particularly not at low airspeeds = during take-off and landing, the most critical phases of the flight. Placing the tail at a lower average incidence angle than the main wing means that there is Angle of Attack margin: when the main wing approaches stall AoA, the tail surface has some AoA left before stalling. This implies that in cruise, if the wing has zero pitch angle, the horizontal tail has negative pitch angle.

But despite the lower incidence, the horizontal tail could stall or partly stall, for instance in a sideslip where a large portion of the horizontal tail is in the wake of the aft fuselage. If it does, we want the aircraft to have a nose down moment, not a nose up moment. If the wing is not stalled yet, the lift force is now not in n.p.$_{fixed}$ anymore, but in a.c.$_W$ And that implies that in the above situation, the nose moves up and the main wing will stall as well. Only when the c.g. is in front of the a.c.$_W$ and the tail force is downwards, will there always be a nose down moment in a stalled tail situation.
And in line with 2. and 3.: it must be certain that there is a stabilising moment in all flight circumstances, at all airspeeds, all angles of attack, all angles of sideslip. That is hard to guarantee when the main wing is optimised for cleanest aerodynamic flow and the stabiliser is in the wake of the main wing: efficiency of the main wing will be higher in all circumstances. By choosing an equilibrium state where the horizontal tail produces a downforce, and therefore less lift with increasing AoA, this problem is solved and the guarantee can be issued, without any additional analysis required. Very good in the olden days of aviation, before CFD.

Note that all above considerations are not really a huge concern in cruise: can the horizontal tailplane ever encounter a vertical wind gust with a velocity that can stall it? The velocity triangle says no, this is not likely to occur in 10$^9$ flying hours. A large aircraft can apply Relaxed Static Stability in cruise: pump fuel to the tail tanks and reduce average down force plus associated trim drag to a minimum, because AoA at cruise is very low. Or trim such that the stabiliser has an upforce, like in the drawing where there is static stability. But at start and landing, the trim tanks must be empty!
The additional burden for certifying this situation is a more elaborate analysis considering gusts at cruise altitude and the influence on static stability: are there any vertical gusts at a higher frequency than 10$^{-9}$ flying hour that can destroy the static stability? Such an analysis was impossible in the starting days of aviation, it is possible now of course.

Anonymous down voter: what is your issue with this answer?

The funny fact is that the tail load is the least negative at  low speed, just when the elevator deflection optimizes the tail airfoil for maximum downforce. At least with positively cambered wings and no flaps, that is. The highest downforce is encountered in a dive at high speed.

Actually, although this is kinda deep into the physics, the weight of the aircraft is not an aerodynamic force and does not play any part in this analysis. We think about it only because we are performing the mathematical calculations in an accelerated frame of reference. The analysis would be identical to that done for an aircraft in a giant box in deep outer space, that had a rocket on it that was generating 32 ft/sec2 linear acceleration, if we performed the calculations in the accelerated  frame of reference of the box. That aircraft has no weight acting on it, it is in free fall.

Just like an aircraft in flight in the earth's atmosphere. It, too is in free fall, cause the force acting on it's wheels when it's parked on the ramp is not there.

another quick point to make, that does not dispute anything in this answer, is that the distinction made between low speed and high speed is totally dependent on the assumption that the aircraft is in a static One "G" flight condition. Although all the principals and factors affecting stability are of course true when at one "G", they must also be identically true at any G-Loading. as well as at any aircraft (bank or pitch) attitude.

Although the answer is not absolutely yes, in almost all conventionally designed (where elevator control is behind the main wing, not a canard) aircraft, the answer is YES, that is the reason.
If the elevator control is behind the main wing, and there is any significant size difference between the main wing and the tail surface, the overall Aerodynamic center will be behind the CG, and the aerodynamic center (the center of pressure) of the main wing (which, subsonicly, is at the 25% of MAC point), will also be behind the CG, not in front of it. Therefore, lift from the main wing will produce a nose down pitching moment, and therefore the tail must be cambered and installed with negative incidence to produce negative (downwards) lift and nose up pitch moment to counteract it.
In order to position the aerodynamic center of the main wing in front of the CG in an aircraft designed with the elevator control surface at the tail, and still have the overall aerodynamic center be behind the CG (necessary for positive static stability), without considering the relative angles of incidence, the relative ratios of the Main wing moment arm to the tail surface moment arm, and the ratio of the lift produced by he main wing to the lift produced by the tail, as depicted in the diagram in @Koyovis' answer would have to satisfy the following inequality.

W = Lift from main (forward) wing
T = Lift from elevator control (Tail)
w = length of Moment arm from main wing AC to CG
t = length of Moment arm from elevator control AC to CG

then, in order for the overall AC to be behind the CG in an aircraft designed with Main wing AC in front of CG, the following would have to be true.
     t/w > W/T

i.e, the ratio of the tail control moment arm to the main wing moment arm must be greater than the ratio of Main wing Lift to Tail Lift. And it must be significantly greater, because the closer to equality these ratios are, the closer to neutral stability the aircraft is. So the inequality must be sufficiently large enough (I confess I don't know by how much that is), but it must be large enough to produce positive stability. To make it larger, you must make w smaller (move the wing AC closer to the CG), or t bigger (move the tail further back), or make the relative sizes of the wing and tail surface more equal (make the tail bigger and the wing smaller). These configurations are possible, and aircraft have been designed like this, but it is the exception, not the most common aircraft design.

Yes the CoG must be in front of the lift neutral point, in all flight circumstances, at side slip angles with half the horizontal tail ineffective etc. What you are describing is that the wing and fuselage have a de-stabilising effect, the tail has a stabilising effect, and the tail stabilising effect must be higher in order to generate a stabilising total aeroplane configuration. That is certainly true, but does not explain why it is more beneficial to use tail downforce than upforce although both situations can have static stability in cruise.

Here I am not sure (perhaps I should post a question to see if someone can address this) but is it possible that this scenario is never used because the band in which the forward AC would have to be is so narrow that forward movement of the AC as AOA increases towards CLmax (Stall AOA) due to wing  trailing edge boundary layer separation could too easily move the overall AC forward of the CG and change positive stability to unstable at the critical point of approach to stall?  https://www.reddit.com/r/aviation/comments/2kibfo/why_does_the_centre_of_pressure_move_rearward/

If this analysis is correct, then it would mean that designing an aircraft this way (where wing AC is forward of CG and both Wing Lift and Tail Lift are positive [upwards] in normal cruise), would be inherently dangerous, as the forward movement of the Center of pressure as AOA increases towards stall would be reducing stability and potentially crossing the threshold and producing an unstable aircraft at approach to stall AOA (like during Landing phase)

There are a lot of explanations on here, but the most simple explanation I can give for the down force of the tail. The downforce of the tail counteracts the rotational moment of lift of the primary airfoil.  This can be overcome with a design using canards.  However, there are drawbacks for this particular design.

Welcome to aviation.SE. While this answer is correct, it does not add details to what is already available in other answers. Would you mind expanding it so to provide more information not already available?


text:
Why was the C-5 considered the largest plane possible?

We started to build the C-5 and wanted to build the biggest thing we could

https://en.wikipedia.org/wiki/Lockheed_C-5_Galaxy
The An-225 is 50% larger and has a shorter life.
Did they mean the C-5 was actually the largest physically possible aircraft or just that anything larger would have weakening service life?
Question reposted, originally posted from an account that had to be deleted.

Interesting side note:
Boeing and Lockheed both bid for the contract to build the C-5 for the US military. Lockheed won. Trying to deliver (almost / did ) bankrupt the company; they merged with Douglas a short time later. Boeing decided not to waste all the engineering studies etc. they had done making their presentation, and used that knowledge to build the 747. The 747 did NOT bankrupt the Boeing company. :)!

Not really. As you suggested, AN-225 is physically bigger than C-5. However at the first flight it is the biggest plane at the time.

Note that "the biggest aircraft we can build" does not mean "the biggest aircraft physically possible." Physics doesn't break down at a certain size. It just gets harder and more expensive, and the advantages don't keep pace.

@TypeIA : unless you get to sizes which can collapse into a singularity, then physics do break down :)  But that's far, far beyond the scope of aircraft engineering.

@Greg The 747 bet did not bankrupt Boeing, but they bet the farm developing it. The Everett plant was built for 747 and future aircraft since Renton was handling 737, 727 and didn't have the runway to add a larger plane.

"biggest thing we could"

There is a "we" in there. And there is a strongly implied "At the time this statement was made in 1963". And there is a very likely addendum of "that would be economically and practically sensible".

@PcMan exactly right.

"The 225 is 50% larger and has a shorter life."...  not a fair comparison, you are comparing a series-production workhorse with a once-off mutant plane..  For a fair comparison of service life, compare the AN-225 with the twin body C-5 Shuttle Carrier. (hint: the c-5 shuttle carrier's lifespan is 0 years, the AN-225 shuttle carrier's lifespan so far is 36 years.

@PcMan Perhaps you would like to add that in an answer?

If I remember correctly, only one AN-225 planes was built, with one started.
For a better comparison (i.e. a plane that is meant for a fleet of them), you might consider the AN-124. It has a similar max takeoff weight (406 metric tons versus 416 for C-5), and roughly similar dimensions

If this is taken from another question shouldn't this be made a community wiki?

@KamiKaze Yes, fair point.

The largest airplane possible is a function of time. At any point in time, the then current technology will have a practical maximum size due to square-cube effects mentioned in this answer:
However, technology advances. The C-5 was developed in the 1960s, with a MTOW of 417 ton. Compared to the 142 ton of the then current 707-320, this was a groundbreaking engineering effort at the time. And no, mounting more engines also runs into problems at the then current technology, like the Spruce Goose experienced, which did not fly out of ground effect.

Did they mean the c 5 was actually the largest physically possible aircraft or just that anything larger would have weakening service life?

Lifespan is indeed one of the design parameters for size, however the C-5M is a C-5 with upgraded engines and avionics according tho the wiki article, so some of the lifespan of the original planes was already consumed. The An-225 with its MTOW of 640 tons was developed in the 1980s.

Your claim that the Hughes H-4 Spruce Goose “could not fly out of ground effect” seems like uninformed speculation. https://aviation.stackexchange.com/questions/12500/was-the-hughes-h-4-hercules-capable-of-flying-out-of-ground-effect

Perhaps "did not fly out of ground effect" would be less ... problematic?

I would reword the first sentence to "The largest airplane possible is a function of ingenuity and available resources."

Can’t the square cube law “easily” be countered with multiple fuselages? Basically just connecting multiple aircraft at the wingtips or stacking them vertically. Though I guess at that point it would be better to fly them as individual aircraft for multiple reasons.

@Michael The stacked aeroplanes need to land, and double deckers went out of fashion a while ago Yes some of the square-cube effects can be overcome to some extent, but the plane needs to be a suitable plane for its mission. Which for almost a century has been a single fuselage with a single wing.

@MikeSowsun If it never actually did, isn”t it speculation to say that it could have?

@koyovis No, it is not speculation to understand that the Spruce Goose on Nov. 2nd 1947 was at a weight considerably less than it’s max design weight and could have easily flown much higher than ground effect.  It had a designed service ceiling of 20,900’.  The only warranted speculation would be whether or not it could operate efficiently at it’s max design weight of 400,000lbs.

Hughes didn't fly it out of ground effect because his only objective was to show that it flies and then put it away, the requirement for it having kind of evaporated.

@Michael Like the He-111Z.... one of the big problems is, with two fuselages side by side, you need even wider runways than with a single big bird...

So if money/development time were no issue, what'd the theoretical limit of today's technology be in terms of wingspan/length/weight?

@Sixtyfive depends on the mission. For passenger aircraft, probably the A380. For cargo aircraft, depending on the cargo density…..The design process is such a multi-dimensional matrix of design goals, refinements, tests, re-designs, that it is not really possible to make a sweeping statement of exact dimensions.

There are other considerations at work here which need to be taken into account.
First of all, neither the An-225 nor the C5 represent the "largest possible plane". If you install enough engines, you could fly a plane larger than either of them, at which point the question becomes where can you land it (maximum runway loading in lbs/sq.ft and wingspan clearance) and where can you shelter it for maintenance. There are few airports in the world that can support the double-decker Airbus because it is too heavy for their runways and their terminal buildings can't load passengers from four skyways into one plane at the same time.
Planes of unusual size tend to be built in support of a specific business plan or use model i.e., transporting objects by air which cannot be transported by ground. Examples include ballistic missiles and the like. But the "divide-by-zero" in this context is that anything you want to put on the plane had to get to the airport by ground in the first place, unless it was built next door to an airport and delivered for use next door to one. These are rare applications for which it is hard to justify constructing a plane from scratch in service of that application which might only be used a hundred times (for emplacing surface-to-surface nukes) from program start to finish.

It may be worth mentioning that the An-225s sole reason to exist was the transport of Buran - just as you said - from its place of construction to the launch site. Similarly, the Mil W-12 (largest helicopter ever) was shrunken to the Mi-26 series version when the size of the Soviet 8K67 intercontinental missiles, for the transportation of which it was intended, could be reduced by the introduction of the new R-29 type.

I beg to differ on the sole purpose of the 225. It may have been the official reason, but why go through all the trouble with the cargo hold if it was only to transport the Buran on its back?

@Jpe61 To also be able to carry the booster engines.

@Davidw no, the An-225 was designed for ferrying ICBMs and other large loads to launch sites, including rocket stages for space launchers. It wasn't just designed for the Buran, though it was adapted for that during its design phase.


text:
Is thrust larger than weight in a vertical climb?
Image source
Very much related to the many questions and great answers to lift and weight in a climb - if an F-16 climbs vertically at a constant speed, is thrust greater than weight?

Comments are not for extended discussion; this conversation has been moved to chat.

Yes, thrust force vector must have a force equal to (but opposite) the combined gravitational force and aerodynamic drag force (from velocity) to climb at constant speed.
In horizontal flight, thrust equals drag to fly at constant speed.  But some of the drag comes from the wing creating lift.  Since the larger wing moves "a lot of air a little", it is more efficient at creating vertical lift than pointing an engine skyward.
That same F-16 therefor needs less thrust to fly horizontally at the same speed.  Climbing vertically, it is essentially an air breathing rocket.
So, since the wing is useless in a vertical climb, the pilot would reduce AOA to zero lift, meaning at the same velocity the vertical plane has less aerodynamic drag than the horizontal one!.
But, thrust required for lift and drag is much less in the horizontal because the wing creates vertical force far more efficiently than the engine.
One can see how this apportionment of thrust to flight requirements leads to a consideration of Drag as Cd0 (drag from velocity) and Cdi (induced drag from lift), perhaps as T0 and Ti.

The first paragraph is very clear.

Another way to look at it is going down the runway before rotation.  Less Cdi, better acceleration.  I like trikes.

The thrust vector is in the opposite direction from the sum of the gravitational and drag vectors, so they are not equal.  At constant velocity, they have equal magnitudes.

Noted and edited, thanks.


text:
Can a commercial pilot with MPL , obtain PPL for general aviation in the UAE?
Is it possible for an airline pilot in case he would like to fly GA aircraft like c152/172 obtain the PPL license without going through the whole process?

Note: You did not specify a country. I will answer for EASA.
You don't need to obtain a PPL, you may use the MPL (Multi-Crew Pilot Licence) if you fulfill the requirements. From EASA Part-FCL (emphasis mine):

FCL.405.A MPL — Privileges
(a) The privileges of the holder of an MPL are to act as co-pilot in an aeroplane required to be operated
  with a co-pilot.
(b) The holder of an MPL may obtain the extra privileges of:
(1) the holder of a PPL(A), provided that the requirements for the PPL(A) specified in Subpart C are met;
(2) a CPL(A), provided that the requirements specified in FCL.325.A are met.
(c) The holder of an MPL shall have the privileges of his/her IR(A) limited to aeroplanes required to be
  operated with a co-pilot. The privileges of the IR(A) may be extended to single-pilot operations in
  aeroplanes, provided that the licence holder has completed the training necessary to act as PIC in
  single-pilot operations exercised solely by reference to instruments and passed the skill test of the
  IR(A) as a single-pilot.

You can find the requirements of Subpart C in the same PDF on page 145f.

Would someone please define "MPL"?  (specifically, the "M"...)

MPL stands for Multi pilot License.

@Bianfable regarding the country lets say it's in United Arab emirates.

@MichaelHall Done.

@IslamAlmaghraby I'm sorry, I don't know about the rules in the UAE. Feel free to edit your question accordingly, maybe someone else will answer.

@Bianfable Thanks.


text:
Can we use the mobile phone GPS at 36,000 ft?
Is it possible to use the built-in GPS function via the smartphone at high altitude during flying as a flight crew ?

GNSS constellation satellites (Navstar, Glonass, Galileo, etc) orbit at 20,000 km altitude, you have some margin. There are unclear legal constraints in altitude and speed set by US COCOM. You need to have a well designed receiver and be near a window if possible, to maximize the signal from the satellites, as what is true in a car is also true in an aircraft (metal blocks signals)

Questions about the experience of passengers on commercial flights (which I assume is what you're asking about here) are off-topic on this site. Also, I'm pretty sure this one has already been answered, either here or on [travel.se].

related, if not duplicate: https://aviation.stackexchange.com/q/42617/1467


text:
Is the well known textbook-formula for net-thrust oversimplified or do I have a misinterpretation?
In order to understand how Jet-Engines works I came across the "famous" formula for net-thrust:
https://en.wikipedia.org/wiki/Turbojet#Net_thrust
$F_{thrust} = Q_{out} \cdot v_{out} - Q_{in}\cdot v_{in}$
From well known physical law about momentum this sound quite reasonable for me.
But now assume an empty cylindrical chamber with nozzle at the end, mounted statically in a wind channel:

Regardless of any details, just by continuity equation we must have
$Q_{out} = Q_{in}$
and
$v_{out} \gt v_{in}$
Nevertheless, this construction would not produce a net propulsion in left direction. It it were true, we just had to put this "empty" construction on a plane...
I was thinking over that quite a time and become already a little crazy.
Where is my mistake? Or is the formula above not the whole truth?
Anyway, I'm not sure if
$v_{in} = v_{air}$
although it sounds reasonably for me.
But even when I take the formula
$F_{thrust} = Q_{out} \cdot v_{out} - Q_{in}\cdot v_{air}$
a left thrust would be produced, whenever
$v_{out} \ge v_{air}$
Is this the key-factor where I'm wrong?

You forgot to add energy inside the duct. The way you draw it, the entrance is blocked by the smaller exit and very little Q will happen. V$_{in}$ is much lower than flight speed, so the picture needs to include what is happening ahead of the duct.

This was intentionally, to make it absurd. I always assumed, that Vin is ~ flight speed and maybe this is part of my problem. In fact it cannot be true, because vin is not zero when the plane is on ground hold. I wonder, however, how the thrust equation can be in form F = Q(out)*v(out) - Q(in)*v(air). Shouldn't it be F = Q(out)*v(out) - Q(in)*v(in) ?

If your v${out}$ is higher than v${in}$ there must be some pressure drop along the duct so the air can accelerate. A positive v${out}$ must have enough pressure to overcome ambient pressure. So where is the much higher entry pressure coming from? This can only be something close to stagnation pressure. Now you have the large frontal area with stagnation pressure working on it – a lot of drag that needs to be included in the model. And stagnation pressure implies v${in}$ = 0, so for any Q to happen your frontal pressure is a bit below stagnation pressure. v$_{out}$ is below flight speed

You forgot to take pressure into account. This formula works assuming your nozzle is well adapted and the exit flow is at atmospheric pressure. In your case the velocity is increased but the pressure drops as stated in Bernoulli's equation.
$$P_0+ 0.5\rho V_0^2 = P_1+ 0.5\rho V_1^2$$
This means that the pressure force on your nozzle will counteract the thrust provided by the fluid acceleration.
See this explanation  your general thrust equation if the pressure are different is
$$\text{Thrust} = (Q * V)_e - (Q * V)_0 + (P_e - P_0) * A_e $$
Given $A_e$ is the exit aera of the nozzle.

I understand, that the pressure on the  right side is not necessarily the same as the pressure on the left side. Bernoulli's law would give the pressure after some calculation. But then I do not understand, why we can assume that the exit flow pressure should be atmospheric pressure when we have a real jet-engine? What is the difference in terms of aerodynamics as compared to the empty case?

The engine is adding some work to the airflow, increasing the pressure internally, thus when applying Bernoulli you can reach atmospheric pressure at a higher velocity on the exit plane.

With this additional term it gives sense. In my simple case I get an under-pressure on the right side, pulling against the thrust from the fluid. However, is atmospheric pressure necessarily reached at the exit plane? Why is the pressure term not given in the formula I have linked to?

Well it's wikipedia... There are often assumption not written in the simplified formula given. Atmospheric is not always reached but the goal is to come as close as possible to it tanks to the nozzle as this would be the most efficient. Supersonic jet engine are the perfect example with the mach diamond in the exhaust proving that the exit pressure is not similar to the surroundings.

Is there a textbook covering those details, without going too far into construction details? I have a book about jet engines, but it has more than 1000 pages and is far to detailed, because I want just a basic understanding on a physical level.

For basic understanding the link to the NASA web site in the answer should be sufficient ? You've got links to the begginer propulsion handbook in the text
https://www.grc.nasa.gov/WWW/k-12/VirtualAero/BottleRocket/airplane/shortp.html

My main mistake might be, that I assumed wrongly, that the inlet speed is the airspeed and therefore the outlet speed must be greater than the air speed in any case. This is not true at all, because air is somehow hindered by the narrowing tube, giving rise to slower speed at he inlet.

Ah, the joy of simplicity.  The diagram is merely drawn backwards.  A good jet has equal pressure at intake and exit.   The exit has more volume and higher velocity than the inlet after compression and addition of heat.  With jets, the fuel also contributes a slight increase in mass.
As always, start with the simplest case, the ram jet, and try to make it work.  A rocket is even easier, there is no air inlet requirement at all, fuel and oxidizer are added to the combustion chamber.
Using the "rocket" analogy, the compressor is holding a "plug" of more dense air over what is essentially a combustion chamber.  Perhaps easier to understand it if one pictures the inlet as an even smaller diameter of liquid oxygen.
We note in this system: the inlet has the higher pressure (equal to the pressure on the walls of the container), and the outlet has the higher velocity.  The larger outlet stream drives the turbine, which drives the compressor.  It is interesting to note the early axial flow Jumo 004 had only has one set of turbine blades driving its compressor.  A two stage turbine design, the Jumo 012, with more than double the thrust, was to be the follow on.

I did not vote on your answer but I can see why some people downwoted it, your not really answering the question ^^...  Furthermore if the jet exhaust is subsonic, you will have a convergent nozzle, so "the exit has more volume and higher velocity" is wrong in that case.

@MaximEck good point about the nozzle.  Comparing the diameter of the narrowest point of the nozzle to the air inlet (narrowest point of the compressor), the nozzle must be wider because T is higher.  The wider diameter of the exit also allows greater torque (on the longer turbine blades), enabling fewer to drive the compressor blades.


text:
What's the reason behind the 787 and 777's "thicker" engine nacelle?
If you look at images of the 777 and 787 engines (both RR & GE), such as these pics
https://upload.wikimedia.org/wikipedia/commons/thumb/e/e4/B787-2139.jpg/220px-B787-2139.jpg
https://www.ainonline.com/sites/default/files/uploads/2016/04/genxenginecopy.jpg
And here's the 777X:
https://media.bizj.us/view/img/11591269/boeing-777x-first-flight-weather-cancelled-11*1200xx5066-2850-0-222.jpg
You'll notice the forward grey part of the nacelle appears "thicker" than seen on other aircraft, here's an A350 engine for example:
https://www.aerospace-technology.com/wp-content/uploads/static-progressive/Airbus%20A380%20TrentXWB.jpg
The grey part of the engine doesn't appear to be painted as "thick" as with the 777 and 787. Same with all other Airbus/Boeing planes, except the two highlighted ones.
Anyone know why this is? I don't think it's necessarily about just engine size either, as the TXWB is bigger than both the engines on the 787, but the grey part isn't as thick.

The cowl is normally designed by the airplane manufacturer or a contracted 3rd party, not the engine maker. The forward silver part gets quite hot when the cowl anti-ice is on. A particular designer elected to use unpainted metal behind the leading edge on that one cowling design. Why? You'd have to ask Boeing.

Modified; Chihaya Sta, CC0, via Wikimedia Commons

Correct terminology: the depth of the highlighted part that you're asking about is the depth of the "lipskin". The deeper it is before the break between the two materials, the longer the flow remains laminar, reducing drag.
The deeper lipskins were harder to manufacture back in the day, and the incremental improvements you've noticed are thanks to manufacturing advancements and in-service validation. Spin forming is making a come back, and there is also the deep draw method, in forming the geometrically-difficult metal parts.

References and further reading:

One of the manufacturers: https://www.collinsaerospace.com/what-we-do/industries/commercial-aviation/aerostructures/nacelle-systems. There's an interactive illustration that will pop out the explanation:

News about another manufacturer being selected as a supplier: https://www.adsadvance.co.uk/gkn-awarded-contract-for-737-max-777x-and-787-components-and-subassemblies.html.

If you like to take a dive into the patents, you now know what terms to use.


text:
Would a System that Detects Catastrophic Oil Leaks in Non-Pressurized General Aviation be Necessary?
I’m currently enrolled in a college course where my group is tasked with designing a product. Our idea is a catastrophic oil leak detection system for non-pressurized general aviation aircraft, proposed by a team member. However, I’m new to aviation and am trying to gauge the actual necessity and feasibility of such a system.
I’ve conducted interviews and sent out a survey, but I’m uncertain if our product idea addresses a real need in the industry. I tried to research this topic online but couldn’t find much information regarding catastrophic oil leaks in general aviation. I’m not sure if this is the right place to ask, as I’ve reached out to others without much success. I apologize if my post is unclear.
Can anyone provide insights into the following?

The frequency and impact of catastrophic oil leaks in non-pressurized aircraft.
Existing solutions or technologies for detecting these leaks.
Any regulatory or safety considerations that may support the need for such a detection system.

I appreciate any information or thoughts you can share.

How is this different from a low oil pressure light? Whether the oil leaked out or the oil pump failed, the result is the same: the engine will cease to run shortly.

The standard oil pressure indicator basically does exactly this, along with providing information that's useful beyond detecting catastrophic leaks. Those are present in basically every aircraft.

A catastrophic oil leak will come from 2 sources: Front crankshaft seal blowing out, or, an oil cooler or cooler supply hose fails, or some other oil line like a turbo supply line. A front crank seal, you'll know it right away because you won't be able to see out the windshield with all the oil on it. That leaves the cooler and hoses. Instead of a leak detector, I'd design a fluid fuse at the engine outlet similar to the hydraulic fuses used on landing gear to prevent loss of hydraulic fluid following a downstream rupture. The fuse would have a cockpit indicator to announce that it's operated.

@user71659 If you have an internal leak under a cowling, you won't get a cockpit indication on pressure until the oil pump pickup is exposed, at which point 90% of the oil will have been dumped overboard. You would expect to see a slow increase in temperature as the oil volume is reduced, but this wouldn't be obvious right away.


text:
Helicopter V speeds explanation
I am reading Cat A heli procedures, and there is Vtoss speed, so what is the difference between Vx(best angle of climb speed) and Vtoss(take off safety speed) ?

Related: How a takeoff decision point (TDP) for vertical takeoff vehicle is defined?

Vx is the best angle of climb speed, with all engines operating. Helicopters can climb vertically in some situations, in which case their airspeed is zero, so Vx is zero. Vx increases above zero for certain loading and density altitude when the helicopter loses the ability to climb vertically.
Vtoss is the minimum airspeed at which a helicopter can maintain a climb if its most critical engine fails. An engine failure below this airspeed will cause the helicopter to begin to lose altitude until its airspeed increases above Vtoss or it hits the ground.


text:
What plane is this? It’s hanging at a bar in Colorado
Consider:

I can’t seem to find out what plane this is. I keep digging online for small twins with twin tips. It has an Air Force emblem on the sign, but I can’t find anything on it.

Did you do a Google image search?

I did my multi rating in one of those. Great aircraft.

If you look up the bar's website you get a little more detail about this aircraft:
Source: Hangar 101 Billiards

The Cessna 310 over the bowling alley is a navy trainer aircraft. It trained some of the finest pilots in the U.S. Navy.

There are more pictures of the aircraft (and others on display) on the website:

Am I the only person who feels like they would be distracted trying to bowl with that thing in the way?

@Michael You and I have very different bowling techniques if that'd be in your way

What is distracting about a hanging airplane? the word hanging makes me uncomfortable.

It is a matter of perspective(?).

@AzorAhai-him- I'd have trouble not hitting the airplane. I'm that bad at bowling.


text:
Why is thrust said to be constant over speed for a jet engine?
I'm just interested in some basic facts regarding flight. On reading the rather "low level" book "Understanding Flight, David F. Anderson, Scott Eberhardt), I recently came up to the following diagram:

Why is thrust approximately constant with speed for a jet engine? Everything gets mixed up in my head as follows:
Thrust is approximately $F_t=\dot m \cdot \left( v_{gas}-v_a\right) $, where the speeds are outlet speed on the nozzle and true airspeed, respectively.
https://en.wikipedia.org/wiki/Turbojet#Net_thrust
So there are really a lot of parameters. What of them are assumed to kept constant in the right diagram and what are changing? Without that information, the diagram seems meaningless for me. For my naïve opinion, a change of thrust is controlled by changing the thrust lever and thereby changing the fuel content within the combustion chamber. Lets say the lever is at its max or a fixed position (whatever this means). Then while increasing speed the factor $v_{gas}-v_a$ gets smaller, assuming the outlet speed is constant.
So to have still the same thrust there are three possibilities':

outlet speed increases
mass stream increases
both of them increase

I can imagine that mass stream increases "somehow" with airspeed (keeping fuel injection constant), so there might be some cancellation of the speed factor. .But is this the whole story? Why is it canceling completely?
On writing this question another question arises:
Considering an ideal engine without losses, total power is
$P_{tot} = \frac{1}{2} \dot m \cdot (v_{gas}^2-v_a^2) = \frac{1}{2} \dot m \cdot (v_{gas}-v_a)(v_{gas}+v_a) = \frac{F_t}{2}\cdot (v_{gas}+v_a)$
Lets say, the power lever is on a fixed position. Is this position an indicator for the total power of the engine? In other words: is total power to the engine constant, when the lever is on a fixed position? I ask, because when assuming constant thrust this formula would mean, that total power (as well as propulsive power) must also increase, because at higher TAS of the plane, gas outlet speed would (up to my poor understanding) rather increase instead of decrease. So what happens, when the thrust lever is on a fixed position? Could it be, that there is a governor behind, taking control of thrust? In this case the right picture has no meaning at all, because while increasing TAS the governor automatically injects more fuel.
There are so many questions and in fact no understanding of anything...

mass flow and outlet speed increasing with airspeed has made sense for me.

see if this helps

yes, it is exactly what I'm asking for.

There are several effects which in combination make constant thrust a good approximation at subsonic speed.
Thrust is created by accelerating a working mass in opposite direction. Net thrust is the difference between the impulse of the air flowing towards the engine and the combined impulse of burnt fuel and the air exiting the engine (and propeller, if one is fitted), derived after the time. Since that impulse is the product of mass and speed, you can either accelerate a large mass by a small speed difference, like a propeller does, or a small mass by a large speed difference, like a turbojet does.
When flying faster, the entry impulse of a propeller quickly grows large relative to the exit impulse, so thrust goes down with the inverse of speed. On the other hand, the high exit speed of a turbojet results only in a small increase of the entry impulse relative to the exit impulse while speed increases.
But if that were all, even the thrust of a turbojet engine would drop when speed increases. But there is a second effect which helps to let thrust grow with speed. With the square of speed, to be precise. That is the ram effect which helps to precompress the air entering the engine. At subsonic speed, this just about compensates for the loss of thrust: At low speed, the growing entry impulse lets thrust drop a bit but at higher subsonic speed the ram effect becomes larger and raises thrust again, such that a constant thrust becomes a good approximation. However, at supersonic speed the ram effect becomes dominant and thrust grows with speed squared – until the absolute internal pressure becomes too high so the engine must be throttled (or the aircraft needs to fly higher) or the shock losses in the intake become too large and thrust drops again.


text:
Does the temperature of the body of an aircraft increase its efficiency?
I am working on a prototype for an airplane. I was wondering, can the temperature of the body of the plane make it more efficient? If the structure is at 25°C, does it need less energy to maintain a steady state flight than a structure at -20°C?
I had this idea about excess energy that could be collected during the day, and was wondering if there is any chance that when the batteries are fully charged and there's still excess power being generated, I could use it in order to increase the performance of the aircraft.

I'm not completely sure what you're asking here. The first part of your question is about the temperature of the aircraft's fuselage (I think), but the second part seems to be about collecting and using solar energy. It might be helpful if you can add some more details about your design, especially if it's electrically powered or not. If yes, this question might be relevant.

For the airspeeds/aircraft you would be powering with a solar/electric system, the temperature of the fuselage would not have any appreciable performance benefit. You would expend a LOT more energy trying to cool the fuselage than you have extra available, plus the added weight of (Peltier?) would really degrade performance.

Typical aircraft structural materials (e.g. aluminum, composites, titanium, etc) don't appreciably change any functional or performance characteristics over the temperature range most aircraft encounter. The only exception I can think of is the SR-71 Blackbird that accounted for the skin heating effect from high-altitude/high-speed flight expanding the fuselage panels.

Most importantly, "when there's still excess power being generated", efficiency is not a concern! Presumably at that point you're flying 100% electrically powered, with batteries charged 100%. And to make things worse, you're optimizing for a rare situation, while you would need to carry the weight of this contraption around all the time when it would be dead weight.

As your speed increases, your temperature will increase due to fluid friction of the air. When metal gets heated, it becomes weaker and more malleable. Think about when a blacksmith is hammering on a piece of iron over a hot forge. Due to this, your Vne should be reduced so your wings don't fall off.
As far as your initial question, if temperature affects efficiency; not really, unless your body gets so hot and you are going so fast as to alter the shape of the aircraft (which could happen, I suppose.)

The SR71 does go so fast that the shape of the aircraft is altered in flight. That's why the skin is corrugated and the fuel tanks leak when it's on the ground. I don't know that this heating and expansion has any impact on the plane's efficiency, though.

@FreeMan talking about the SR71 in the current tense is a bit weird, considering it was last in service almost 20 years ago.

@Jamiec, that's true, but I'm fairly certain that if they were to put one back in the skies, it would still do all the things it used to do. ;)  Sad to think that it's 20 years gone...

@Jamiec Does NASA not use them any-more for science purposes?

@alex No I dont think so NASA operated the two last airworthy Blackbirds until 1999 - source: wikipedia.

@Jamiec That's a shame.

Even NASA couldn't afford to keep the SR71 in the air. The very expensive JP7 fuel, very expensive oil for the engines, both to withstand the high temperatures the aircraft experiences, and very expensive parts.


text:
Can I become a Commercial Pilot if I had OCD?
In a few weeks, I am going to get off my medication for OCD treatment. The FAA says if episodes are severe enough to cause disruption of vocational or educational activity then I should be disqualified. In the very beginning that was the case, but now school helps me not to think about it.
The FAA says if I am on medication or have suicidal ideation then I should be disqualified. Well, considering I am getting off my medication in a few weeks that means it's not a required medication, right? And I have never had any suicidal thoughts.

Mental health questions are tough to answer because there are so many variables and the human mind is a tricky thing. You will need to talk to an aviation medical examiner for this one to get a definitive answer.

Just to clarify the point about speaking to an AME: ask them for a consultation only, do not apply for an actual medical. If the AME starts a formal medical, it can only end in pass or fail (or deferral to the FAA) and if you fail that can make a lot of things a lot more difficult. So make sure you get everything worked out before you apply.

I think you are required to have OCD to be a pilot.

Maybe. It depends on how severe the OCD was, the root causes of the disorder, the length of time since the last episode and the likelihood of future recurrence. The aviation medical system is set up to be conservative for good reasons, the cockpit is a bad place to become fixated.
If your condition is/was stress related then I would expect it to be a no, flying can be very stressful at times, you have to be able to handle fluid, difficult scenarios.
Your best answer will be from an Aviation Medical Examiner, I suggest you speak to one, it's what they're for.


text:
How long should I wait after I've gotten off my OCD medication to consult an aviation medical examiner?
Im getting off my OCD medicine in a few weeks and wanted to know how long I should wait til i should consult a aviation medical examiner, any advice on which aviation medical examiner I should consult?

Timing probably matters very little as far as talking to the AME. He'll need to consider not just the medication you aren't on any longer (and he can tell you what timeframe has been "long enough" there), but also the nature of your condition without the medication. The time needed to evaluate how you're doing without the medication is probably longer than the timeframe required to "have the medicine out of your system before you go fly."
In cases like cold medicine, the latter is some multiple of the dose interval, and it's pretty clear that the initial condition (the cold) is gone & won't cause you any further problems. I am not an AME, but my expectation would be that OCD is enough different that the timeframe will be much longer.
It will take an AME to tell you what the path from where you are to where you want to be (having your medical issued) looks like and how long it will take.  There are too many variables in these sorts of things for much of a "one size fits all" answer.
As far as which AME, I'd suggest calling a few in your area & asking if they have experience with patients with OCD, and if not, whom they'd recommend.  It may take a bit of research, but it should be possible to find one who had done this before.  Ideally, "local" is better than far distant, although if may be possible for the right AME, even if he's some distance away, to work with your local doctor to gather what's needed so he can send it to the FAA & get approval, so you might only need 1 in-person visit with him.  I'd definitely want somebody familiar with more than the routine "check eyes & BP and sign the paperwork" procedures!


text:
Tiller use on takeoff
When a commercial jet aircraft has been cleared for takeoff, does the PF use the tiller for steering, up to a certain speed? I've seen many vids where it seems the PF uses the yoke/sidestick for steering from the very beginning of the takeoff roll.

See also https://aviation.stackexchange.com/questions/50877/when-does-an-airliner-switch-from-using-the-tiller-to-the-rudder

Usually, the nosewheel steering provided through the rudder pedals (not the yoke or sidestick) is sufficient. The tiller is useful for large deflections of the nosewheel, like turning from one taxiway to another that's 90 degrees off, but for small corrections like tracking the centerline, the less sensitivity & lower authority of the rudder pedals is preferable.
On takeoff roll, I typically guard the tiller until 80 knots, mainly in case of a sudden engine failure (or, worse, an engine seizure), which could try to yank the aircraft off the side of the runway pretty fast.  Above 80 knots, you have the authority of the rudder itself at work, and the tiller would be too sensitive even in the the engine seizure scenario. So above 80 knots you really don't want to be touching the tiller.
The case where you wouldn't have nosewheel steering thru the rudder pedals would be if the interconnect from the pedals to the NWS was deferred; I know it's in the MEL, but I've never seen it.  In that case, you'd need at least some pressure on the tiller to keep things tracking straight until there is enough airflow on the rudder that the pedals have adequate authority. And, in the case of an engine failure, you'd be going to the tiller right away anyhow -- and pulling the operating engine back to idle, and on the brakes (differentially, if needed), so there should be plenty of ways to maintain directional control in an engine-out scenario, even with that deferral.
In the C-130, there was no interconnect between the pedals and the NWS, so everything was the tiller, and sometimes differential power (or reverse) and differential braking (rarely). Instant response from the props made that a very viable way to taxi.  And that worked fine; the rudder became effective pretty quickly on takeoff roll -- having lots of airflow coming off the props probably helped out.
But the normal case in a 737 is, the pilots steer the nosewheel via the rudder pedals for the entire takeoff roll, and the tiller isn't used, only guarded.

I've tried no NSW in the CRJ sim using differential braking. Differential power to steer was less effective with the tail mounted engines. Sharp jabs of brake were necessary.  The B-29 had a free castering nosewheel and was reportedly a handful on the ground at the middling speeds below rudder effectiveness and was heavily dependent on differential outboard engine power.


text:
Ram Rise and Total Air Temperature explantation
Since ram rise is the same as total air temperature, why does Total Air Temperature minus Static Air Temperature give ram rise? What is really the meaning of ram rise: is it the kinetic components acting on an aircraft, or what?

Does this question and its answer give you what you're looking for?

Hint: the opening statement "Since ram rise is the same as total air temperature..." is incorrect.  The addition you mention, SAT+RR = TAT, is explained in the answer I linked, and that's what you should focus on there.


text:
What's prohibiting the use of large batteries for electric aircraft?
In our society, the majority of the aircraft are powered by fuel. Recently, electric aircraft using electric batteries have begun to appear.
However, the range of these aircraft are limited by the amount of energy in its batteries.
So the question has to be asked; can't we just add larger batteries for extra energy? I understand that large batteries increases the weight of the aircraft, but certainly the energy provided can generate lift far greater than the weight added by the size enlargement, right?
Some articles regarding the battery issue:
https://theconversation.com/climate-explained-why-dont-we-have-electric-aircraft-123910
https://sbi.sydney.edu.au/why-dont-we-have-electric-aircraft/

Both answers are good and very similar to the other "why don't we have electric commercial airliners yet" questions.  This is definitely a dupe, just don't have time to find it right now.

"...but certainly the energy provided can generate lift far greater than the weight added by the size enlargement, right?" Wrong!

Gosh, just add bigger and more batteries.  Why didn’t the engineers think of that?!

I don't quite understand the downvotes...

@jamesqf It's not wrong at all. It's only ignoring the more interesting question that is: "for how long?"

"Ok, so we dig an escape tunnel, but where are we going to hide the excavated earth?"...."Easy, we dig another tunnel and bury it".

Please do not vandalize your question by deleting the text. Under certain conditions, you can delete your own question, but realize that those who responded put time into writing answers for you (and in this case, I don't think the question can be deleted now). Please be respectful of their work by not removing the question to which they responded. It's closed, but still visible, and can be found in searches to help point those with a similar question to the best answers.

Jet fuel is around 43 MJ/kg while battery is 0.9MJ/kg
Fossil fuel airplanes is already taking off with 1/3 total weight being fuel 1/3 weight being payload (B737-300, 62t MTOW, 18t fuel, 16t payload). On long trips fuel weight goes up to about half MTOW, so you are facing some choices:

Same MTOW and payload weight, electric have about 4% of the range of fossil fuel (electricity is 1/50 as energy-dense but 2x efficient).
Same MTOW, no payload, electric have 8%-12% of the range.

And point 2 answers your question: no, power doesn't solve your weight issue. Even an airliner fully loaded with batteries to MTOW doesn't fly more than 500 miles. But in practice even loading a B737 with batteries is too expensive to even build a demo.

But even with commercial planes, the weight of the fuel doesn't STAY at 1/3 to 1/2 the takeoff weight, but declines throughout the flight.  Batteries always weigh the same.

Yes, @jamesqf. Until we find a way to jettison the electrons right off the plane after they have done their work having moved from one battery pole through the engine and then back to the other pole. I mean, such an electron is quite heavy, ain't it? And there's a whole flock of them. ;-)